Maximizing Area: Among all rectangles that have a perimeter of 20 ft, find the dimensions of the one with the largest area.
To find the dimensions of the rectangle with the largest area among all rectangles with a perimeter of 20 ft, we need to use the concept of optimization. Let's break down the problem into smaller steps:
Step 1: Set up the problem
Let's assume the length of the rectangle is 'l' feet and the width is 'w' feet. We are given that the perimeter of the rectangle is 20 ft, so we can write the equation:
2l + 2w = 20
Step 2: Solve the equation for one variable
Let's solve this equation for 'l'. Subtract 2w from both sides:
2l = 20 - 2w
Divide both sides by 2:
l = 10 - w
Step 3: Substitute the expression into the area formula
The formula for the area of a rectangle is given by A = length * width. Substitute the expression for 'l' into the formula:
A = (10 - w) * w
A = 10w - w^2
Step 4: Maximize the area
To maximize the area, we need to find the value of 'w' that gives the maximum value for 'A'. We can do this by finding the vertex of the quadratic function A = 10w - w^2.
Step 5: Find the vertex of the quadratic
The vertex of a quadratic function y = ax^2 + bx + c is given by the formula x = -b / (2a). In our case, a = -1, b = 10, and c = 0 because there is no constant term.
w = -10 / (2*(-1))
w = -10 / (-2)
w = 5
Step 6: Find the corresponding length
Use the value of 'w' to find the corresponding length. Substitute w = 5 into the equation l = 10 - w:
l = 10 - 5
l = 5
So, the rectangle with the largest area among all rectangles with a perimeter of 20 ft has dimensions of 5 ft by 5 ft.